php - Android 向 PHP 发送字符串

Question

我正在尝试使用 php 将 android 与 mysql 连接起来，将参数包含到表中。

我编写了可以运行的 php 代码。但我不明白为什么 android 的代码不起作用。我在 logcat 中没有收到错误。

主要 Activity :

...
boton.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View v) {

                sendMessage("mensaje");
            }
        });
    }


    public void sendMessage(String t){

        ServerRequests serverRequests = new ServerRequests(this);
        serverRequests.fetchUserDataAsyncTask(t);{
            Log.d("", "sendMessage: "+t);
        };
    }
...

服务器请求

> public class ServerRequests {
> 
>     URL url;
>     HttpURLConnection conn;
>     ProgressDialog progressDialog;
>     String mensaje;
> 
>     public ServerRequests(Context context) {
>         progressDialog = new ProgressDialog(context);
>         progressDialog.setCancelable(false);
>         progressDialog.setTitle("Authenticating...");
>         progressDialog.setMessage("Please wait...");
>     }
> 
>     public void fetchUserDataAsyncTask(String mensaje) {
>         progressDialog.show();
>         new fetchUserDataAsyncTask(mensaje,"hola").execute();
>     }
> 
>     public class fetchUserDataAsyncTask extends AsyncTask<Void, Void, String> {
> 
>         String returnedUser;
> 
> 
>         public fetchUserDataAsyncTask(String mensaje, String returnedUser) {
> 
>         }
> 
>         @Override
>         protected String doInBackground(Void... params) {
> 
>             try {
> 
>                 url = new URL("http://gclimb.com/androidphp/index.php");
> 
>                 conn = (HttpURLConnection) url.openConnection();
>                 String param = "mensaje="+mensaje;
>                 conn.setRequestMethod("POST");
>                 conn.setDoInput(true);
>                 conn.setDoOutput(true);
>                 conn.setRequestProperty("Content-Type", "application/x-www-form-urlencoded");
>                 conn.setRequestProperty("charset", "UTF-8");
>                 conn.connect();
> 
>                 OutputStreamWriter out = new OutputStreamWriter(conn.getOutputStream());
>                 out.write(param);
> 
>                 //No code to receive response yet, want to get the POST working first.
> 
>             }
>             catch (MalformedURLException e) {
>                 e.printStackTrace();
>             } catch (ProtocolException e) {
>                 e.printStackTrace();
>             } catch (IOException e) {
>                 e.printStackTrace();
>             }
>             catch (Exception e) {
> 
>             } finally {
>                 progressDialog.dismiss();
>             }
> 
>             return returnedUser;
>         }
>     }
> 
> }

编辑

这是简单的index.php 文件。

php 文件:

<?php

$mensaje = $_POST['username'];
$mensaje2 = $_POST['password'];
$bo = $mensaje;
$servername = "zzzzzzz";
$username = "zzzzzz";
$password = "zzzzz";
$dbname = "zzzzz";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

$sql = "INSERT INTO AndroidPhp (mensaje) VALUES ('$bo')";
$conn->query($sql);





$sql2="SELECT mensaje FROM AndroidPhp";

$result = $conn->query($sql2);

if ($result->num_rows > 0) {
    // output data of each row
    while($row = $result->fetch_assoc()) {
      echo $row["mensaje"];
    }
} else {
    echo "0 results";
}

 ?>

Answer 1

使用 Volley 库。很简单。

Samples : Volley Library

您可以找到my answer也是。